I understand (1)-(2) in the following way: for every object $i\in I$ there exists a pair $(f_i,g_i)$, such that $f_i\colon i_0\to i$ and $g_i$ is its left inverse. Am I right?
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OskarDec 5 '13 at 18:45

2 Answers
2

No: the conditions are trivially satisfied for any filtered category with one object, e.g. the category freely generated by an idempotent endomorphism.

That said, it is almost true. Consider the full subcategory of $\mathcal{I}$ spanned by $i_0$, and let $P$ be the inclusion. I claim $P$ is cofinal, i.e. that for all $i$ in $\mathcal{I}$, the comma category $(i \downarrow P)$ is connected. Indeed, $(i \downarrow P)$ is inhabited, by condition (2), and given any two morphisms $g, g' : i \to i_0$, there is a pair of morphisms $h, h' : i_0 \to j$ such that $h \circ g = h' \circ g'$ (by filteredness), and then $g_j \circ h \circ g = g_j \circ h' \circ g'$, therefore $g$ and $g'$ are connected by a cospan in $(i \downarrow P)$. Note that we do not need condition (1).

It is well-known that replacing a diagram with a cofinal one does not change its colimit, i.e. $\varinjlim F \cong \varinjlim F P$. Thus we can replace $\mathcal{I}$ with a category that has only one object (but possibly more than one morphism).

No. Let $\mathbf{Set}_{0,1}$ denote the full subcategory of $\mathbf{Set}$ without empty set and all singletons(such category enjoy requirements (1)-(3), and $i_0\cong\{x,y\}$), and $V\colon\mathbf{Set}_{0,1}\to\mathbf{Set}$ be an inclusion functor. The colimit of $V$ is a singletone, not $V(\{x,y\})$.

Actually, this works for many subcategories of $\mathbf{Set}$, even for $\mathbf{Set}_{\{x,y\}}$(full subcategory of $\mathbf{Set}$ with one object - fixed two-element set).